题目内容
如图,已知AC平分∠PAQ,点B、B′分别在边AP、AQ上,如果添加一个条件,即可推出AB=A B′,那么该条件不可以是( )
A.BB′⊥AC B.CB=CB′ C.∠ACB=∠ACB′ D.∠ABC=∠AB′C
A.BB′⊥AC B.CB=CB′ C.∠ACB=∠ACB′ D.∠ABC=∠AB′C
B.
试题分析:如图:∵AC平分∠PAQ,点B,B′分别在边AP,AQ上,
A:若BB′⊥AC,在△ABC与△AB′C中,∠BAC=∠B′AC,AC=AC,∠ACB=∠ACB′,∴△ABC≌△AB′C,∴AB=AB′;
B:若BC=B′C,不能证明△ABC≌△AB′C,即不能证明AB=AB′;
C:若∠ACB=∠ACB′,则在△ABC与△AB'C中,∠BAC=∠B′AC,AC=AC,∴△ABC≌△AB′C,∴AB=AB′;
D:若∠ABC=∠AB′C,则∠ACB=∠ACB′∠BAC=∠B′AC,AC=AC,∴△ABC≌△AB′C,∴AB=AB′.
故选B.
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